2013 HSC Mathematics Extension 1 Marking guidelines

2013 HSC Mathematics Extension 1 Marking Guidelines

Section I

Multiple-choice Answer Key

Question Answer 1 C 2 D 3 C 4 D 5 A 6 B 7 A 8 D 9 B 10 C

– 1 –


Section II

Question 11 (a)

Criteria Marks • Correct answer 1

dFor any cubic αβγ = − , so

a 7αβγ = − . 2

Question 11 (b)

Criteria Marks • Correct primitive 2 • Attempts to use standard integral, or equivalent merit 1

⌠ 1

49 − 4x2 dx

⌡⎮

⌠ ⎮ ⎮⎮⌡

1 dx =

⎛ ⎝⎜

⎞ ⎠⎟

49 2− x4 4

⌠ ⎮ ⎮⎮⌡

1 2

1 dx =

⎛ ⎝⎜

⎞ ⎠⎟

27 2− x 2

1 x = sin−1 + c 2 7

2

= 1

sin−1 2x + c 2 7

– 2 –


Question 11 (c)

Criteria Marks • Correct answer 2 • Identifies correct binomial coefficient, or equivalent merit 1

The probability is

⎛ 1 ⎞ 7 ⎛ 3⎞ 3 10! 33

10 C7 ⎝⎜ ⎠⎟ ⎝⎜ ⎠⎟ 410 =

4 4 7!3!

33

= 120 ⋅ 410

33

= 30 ⋅ . 49

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝

1For each question, the probability of choosing the correct option is and the

4 3

probability of choosing an incorrect option is .

⎞ ⎠

4

⎛ ⎝

The probability of choosing the correct option for exactly 7 out of 10 questions is

⎜ ⎟ ⎜ ⎟ ⎞ ⎠

⎛ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠

10 71 3 1

the term in + containing . 4 4 4

Question 11 (d) (i)

Criteria Marks • Correct solution 2

• Finds 1

Domain: all real x ≠ ±2

Using the quotient rule

(4 − x )⋅1 − x(−2x)ƒ ′ x( ) = 2

2 )2(4 − x

4 − x2 + 2x2 =

2 )2(4 − x

24 + x = 2 )2(4 − x

ƒ ′( ) x > 0 since 4 + x2 > 0 for all x, and (4 − x2 )2 > 0 for all x ≠ ±2

Hence ƒ ′( ) > 0 for all x ≠ ±2. x

– 3 –


Question 11 (d) (ii)

Criteria Marks • Correct graph 2 • Finds both vertical asymptotes, or equivalent merit 1

x–2 20

y

y x= ƒ ( )

Question 11 (e)


x xsin sin 12 2lim = lim

x→0 3x x→0 6 x 2

xsin 1 2= lim

6 x→0 x 2

1 ⎛ sin t ⎞ = since lim = 1.

6 ⎝⎜ t→0 t ⎠⎟

– 4 –


Question 11 (f)


• Finds a primitive of the form of a tan−1 u 2

• Attempts to use given substitution 1

du 3x 3x 2 6xu = e , = 3e , u = edx

If x = 0, then u = e0 = 1 1

1 3 ⋅ If x = , then u = e 3 = e

3 Hence

1

⌠ 3 e3x

dx ⎮⌡0 e6x + 1

e1 ⌠ 1 = du ⎮3 ⌡1 u2 + 1

e1 = ⎡tan−1 u⎤3 ⎣ ⎦1

1 = 3 ( tan−1 e − tan−11 )

1 ⎛ π ⎞ = tan−1 e −

3 ⎝⎜ 4 ⎠⎟ .

Question 11 (g)


• Uses correct derivative of , or equivalent merit 1

Using the product rule

ƒ ′ x x 5( ) = 2

2 + 2x sin−1 5x

1 − 25x

25x = + 2x sin−1 5x. 21 − 25x

– 5 –


Question 12 (a) (i)


We want

3 cos x − sin x = 2 cos (x + α ) 3

cos x − 1

sin x = cos (x + α )2 2

= cos x cosα − sin x sinα ,

3 1 1So cosα = and sinα = . Hence tanα =

2 2

π ⎛ π π ⎞∴ α = and 0 < <

6 ⎝⎜ 6 2 ⎠⎟

Question 12 (a) (ii)

3 .


• Obtains an equation of the form , or equivalent merit 1

Rearrange the equation and use part (i):

⎛ π ⎞3 cos x − sin x = 1, ie 2 cos x + = 1⎝⎜ 6 ⎠⎟

⎛ π ⎞ 1 cos x + = ⎝⎜ 6 ⎠⎟ 2

π π π 5π∴ x + = or x + = 6 3 6 3

Hence the solutions satisfying 0 < x < 2π are

π π π x = − =

3 6 6 5π π 3π

and x = − = . 3 6 2

– 6 –


Question 12 (b)

Criteria Marks • Correct solution 3 • Finds correct primitive, or equivalent merit 2 • Finds correct integrand, or equivalent merit 1

y

x3π 2

O

Volume of solid is given by

V = π y2 dx ∫ 3π

⌠ 2 ⎛ x ⎞ 2

V = π⎮ ⎝⎜ 3 sin ⎠⎟ dx 2⌡0

3π

⌠ 2 = 9π⎮ sin2 x dx

⌡ 20 3π

⌠ 2 1V = 9π⎮ (1 − cos x)dx

⌡0 2

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝

Note: cos x = cos2 x − sin2 x 2 2

= 1 − 2 sin2 x 2

∴ sin2 x = 1 (1 − cos x)

2 2

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠

3π9π = ⎡⎣(x − sin x)⎤⎦0 2

2

9π ⎡⎛ 3π 3π ⎞ ⎤ = ⎢⎝⎜ − sin ⎠⎟ − (0 − sin 0 )⎥2 ⎣ 2 2 ⎦

9π ⎛ 3π ⎞ = + 1

2 ⎝⎜ 2 ⎠⎟

9π ⎛ 3π ⎞The volume is ⎝⎜ + 1⎠⎟ units3 .

2 2

– 7 –


Question 12 (c)

Criteria Marks • Correct solution 3 • Finds values for A, B and k, or equivalent merit 2 • Finds value for A, or equivalent merit 1

As t → ∞, T approaches the temperature of the room, and so A = 22.

When t = 0, T = 22 + B

= 80

∴ B = 58.

Hence −kt T = 22 + 58e

To find k : After 10 minutes the temperature is 60° (t = 10, T = 60) −k ×10 60 = 22 + 58e

−10 k38 = 58e

⎛ 38 ⎞log = −10 k ,e ⎝⎜ 58 ⎠⎟

−1 ⎛ 38 ⎞ so k = loge ⎝⎜ ⎠⎟10 58

To determine time to drop to 40°C: 1 ⎛ 38 ⎞

log ⎠⎟ t 10 e ⎝⎜ 58 40 = 22 + 58e 1 ⎛ 38 ⎞

18 log ⎝⎜ ⎠⎟ t 10 e 58 = e 58 18 1 ⎛ 38 ⎞

log = log ⎠⎟ t e e ⎝⎜58 10 58 ⎛ 18 ⎞

loge ⎝⎜ 58 ⎠⎟ t = 10 ,⎛ 38 ⎞

loge ⎝⎜ 58 ⎠⎟

t = 27.67...

∴ temperature drops to 40° after 28 minutes (to the nearest minute).

– 8 –


Question 12 (d) (i)


• Uses an appropriate method to attempt to find 1

Using the perpendicular distance formula from P t( ,t2 + 3) to the line y = 2x − 1 :

2t − (t2 + 3) − 1D(t) =

− t2 + 2t − 4 =

4 + 1

5

− (t2 − 2t + 4)=

5

t2 − 2t + 4 =

5

Now t2 − 2t + 4 = (t − 1)2 + 3 > 0 for all t, so

t2 − 2t + 4D(t) = is the perpendicular distance.

5

Question 12 (d) (ii)


To find minimum distance, D′(t) = 0.

D′(t) = 1 (2t − 2) 5

= 2 (t − 1) 5

= 0

∴ t = 1

Minimum because D(t) → ∞ as t → ±∞.

1 )2Alternative: D(t) is a quadratic with minimum at t = 1 since D(t) = ((t − 1 + 3). 5

– 9 –


Question 12 (d) (iii)


t = 1 corresponds to x = 1, on the curve given by y = x2 + 3.

y′ = 2x, so the slope of the tangent at x = 1 is 2.

Also, the slope of the line y = 2x − 1 is 2.

∴ the line and the tangent are parallel, since their gradients are equal.

Question 12 (e)


• Correctly differentiates v2 with respect to x, or equivalent merit 1

v2 + 9x2 = k 2 2v = k − 9x

1 1 92 2v = k − x 2 2 2

d ⎛ 1 2 ⎞ x = v = −9x dx ⎝⎜ 2 ⎠⎟

2For SHM, x = −n (x − b). Now, x = −9x, is of this form with

n = 3 and b = 0

2πAs n = 3, the period is .

3

– 10 –


Question 13 (a) (i)


2Using V = 4 πr3 and A = 4πr ,3

dV −10−4 A = dt dV dr = dr dt

2 dr = 4πr dt

dr = A dt

Hence dr = −10−4 , which is constant. dt


Criteria Marks • Correct answer 2 • Finds initial radius, or equivalent merit 1

From part (i) dr = 10−4

dt To determine the initial radius of the raindrop:

34 πr = 10−6 ,3

3 so r3 = 10−6

4π

33r = 10−2

4π

1 33= 102 4π

Time to evaporate;

r t =

10−4

33= 102

4π ∴ time = 62 seconds (to the nearest second).

– 11 –


Question 13 (b) (i)

Criteria Marks • Correct solution 2 • Writes an unsimplified expression for the x- or y- coordinate of G, or

equivalent merit 1

The point G divides NT in the ratio 2:1 externally means that T is the midpoint of NG.

Hence if (x0 , y0 ) are the coordinates of G, then

0 + x0 = ap (x-coordinate of T )2

2a + ap2 + y0 = 0 (y-coordinate of T )2

2∴ x0 = 2ap and y0 = −2a − ap

ie G is the point (2ap, − 2a − ap2 ), as required

– 12 –


Question 13 (b) (ii)

Criteria Marks • Correct solution for G 2 • Shows that G lies on a parabola, or equivalent merit 1

The original parabola has focal length a and directrix y = −a. To show 2that G (2ap,−2a − ap)2 lies on a parabola, eliminate p from x = 2ap, y = −2a − ap

xNow, p = , so

2a

⎛ x ⎞ 2

y = −2a − a⎝⎜ 2a ⎠⎟

2 y = −2a −

ax 24a

2 = −2a −

x 4a

2 y + 2a = −

x 4a

2x = −4a( y + 2a) This is a parabola with focal length a and vertex (0, – 2a).

So the directrix is

y = −2a + a

= −a

Hence focal length and directrix are the same for both parabolas.

Alternative: 2(2ap,−ap ) is the parabola obtained if the original parabola is reflected

about the x-axis. It has focal length a and directrix y = +a. 2Now G (2ap,−2a − ap ) is obtained by translating the reflected

parabola down by 2a units. Translation preserves the focal length a, and

the directrix becomes y = a − 2a = −a.

– 13 –


Question 13 (c) (i)

Criteria Marks • Correct solution 2 • Obtains an expression for , or equivalent merit 1

We need to find the time when the quadratic y = ut sinα − g

t2 is a maximum. 2

dy = usinα − gt dt

= 0 for max/min

usinα t =

g

This will be at the maximum height since y = ut sinα − g

t2 is a concave down parabola. 2

Question 13 (c) (ii)


Similarly from part (i), the projectile fired from B reaches its maximum height when

wsin β t = .

g

The projectiles collide when they both reach their maximum height.

ie usinα = t =

wsin β ,

g g

so usinα = wsin β.

– 14 –


Question 13 (c) (iii)

Criteria Marks • Correct solution 2 • Finds the distance one projectile has travelled 1

When the projectiles collide, the sum of the two distances travelled must be d. Both

projectiles have travelled for time t.

usinα wsin β t = = , so

g g

usinα wsin βd = u cosα + w cosβ

g g

wsin β usinα = u cosα + w cosβ g g

uw = (sin β cosα + sinα cosβ )g

uw = sin (α + β ). g

– 15 –


Question 13 (d)


• Proves that ∠QBT and are equal AND uses the properties of the common tangent, or equivalent merit

2

• Proves that ∠QBT and are equal or uses the properties of the common tangent, or equivalent merit

1

The circles C1 and C2 have a common tangent at T .

Choose points X and Y on that tangent as marked on the diagram above.

∠YTP = ∠PAT = α (angle in alternate segment, chord PT and tangent XY )

∠XTQ = ∠QBT = β (angle in alternate segment, chord QT and tangent XY )

But ∠PAT = ∠QBT (alternate angles equal, BQ PA)

ie α = β

∴ ∠YTP = ∠XTQ

Hence QP is a straight line making equal vertically opposite angles with line XY at T .

∴ P,T ,Q are collinear.

– 16 –


Question 14 (a) (i)


1 1 1 1 k + 1 k− + = − ( ) + ( ))2 k k + 1 )2 k k + 1 k k + 1(k + 1 (k + 1

1 ⎡ 1 1 ⎤ = k + 1 ⎣⎢ k + 1

− k ⎦⎥

< 0 (since k + 1 > k and k > 0).



• Uses correct assumption in an attempt to prove that 2

• Establishes initial case, or equivalent merit 1

1 1 1 5 1 3Case n = 2 : LHS + = 1 + = RHS: 2 − =

1 22 4 4 2 2 LHS < RHS

Hence true for n = 2

Assume true for some k ≥ 2 k 1 1

ie < 2 −∑ p2 k p=1

Need to show true for n = k + 1 k +1 1 1

ie < 2 −∑ p2 k + 1 p=1

Now k +1 k1 1 1

LHS = ∑ = ∑ +2 2 )2

p=1 p p=1 p (k + 1 1 1

LHS < 2 − + ��(by induction assumption) k )2(k + 1

⎛ ⎞1 1 1 1 1 1< 2 − + − ⎜ using part (i), ie < − ⎟k k k + 1 )2 k k + 1⎝ (k + 1 ⎠ k +1 1 1

ie < 2 −∑ p2 k + 1 p=1

By the principle of Mathematical Induction, it is true for all n ≥ 2.

– 17 –


Question 14 (b) (i)


From the binomial theorem 4n

(1 + x)4n = ∑ ⎛ 4n⎞

x4n−k

⎝⎜ k ⎠⎟ k =0

2n ⎛ 4n⎞ The coefficient of x is

⎠⎟ , choosing k = 2n.

⎝⎜ 2n

Question 14 (b) (ii)


• Writes as or uses the binomial expansion,

or equivalent merit 1

From the binomial theorem

)2n )2n(1 + x2 + 2x = (1 + x(x + 2) 2n ⎛ 2n⎞ )2n−k

= ∑ (x(x + 2)⎝⎜ k ⎠⎟ k =0

2n ⎛ 2n⎞ 2n−k ( )2n−k = ∑ x + 2 . ⎠⎟

x ⎝⎜ kk =0

– 18 –


Question 14 (b) (iii)

Criteria Marks • Correct solution 3 • Substitutes the given expression into part (ii) and makes progress towards

the solution 2

• Connects and , or equivalent merit 1

)2(1 + x2 + 2x) = (1 + x , so

)2n(1 + x2 + 2x = (1 + x)4n

By part (i), ⎛ 4n⎞

⎠⎟ is the coefficient of x2n in (1 + x2 + 2x)2n

. ⎝⎜ 2n

From part (ii) these are

⎛ 2n⎞ )2n−k2n−k 2nthe terms in (x + 2 for k = 0, …, n. If k > n there is no term involving x . ⎠⎟

x ⎝⎜ k

Using the given identity the coefficients of x2n are:

⎛ 2n⎞ ⎛ 2n − 0⎞ k = 0 :

⎠⎟ 22n−0

⎝⎜ 0 ⎠⎟ ⎝⎜ 0

⎛ 2n⎞ ⎛ 2n − 1⎞ k = 1 :

⎠⎟ 22n−1−1

⎝⎜ 1 ⎠⎟ ⎝⎜ 1

⎛ 2n⎞ ⎛ 2n − 2⎞ k = 2 :

⎠⎟ 22n−2−2

⎝⎜ 2 ⎠⎟ ⎝⎜ 2

⎛ 2n⎞ ⎛ 2n − n⎞ k = n :

⎠⎟ 22n−n−n

⎝⎜ ⎠⎟ ⎝⎜n n

Hence

⎛ 4n⎞ ⎛ 2n⎞ ⎛ 2n⎞ ⎛ 2n⎞ ⎛ 2n − 1⎞ ⎛ 2n⎞ ⎛ n⎞ =

⎠⎟ 22n +

⎠⎟ 22n−2 + … +

⎠⎟ 20

⎝⎜ 2n⎠⎟ ⎝⎜ 0 ⎠⎟ ⎝⎜ 0 ⎝⎜ 1 ⎠⎟ ⎝⎜ 1 ⎝⎜ n ⎠⎟ ⎝⎜ n n ⎛ 2n⎞ ⎛ 2n − k⎞

= ∑ ⎠⎟ 22n−2k .

⎝⎜ k ⎠⎟ ⎝⎜ kk =0

– 19 –


Question 14 (c) (i)

Criteria Marks • Correct solution 2 • Correctly applies Newton’s method 1

1Take ƒ ( ) t = et −

t

1Then ƒ ′( ) t = et +

t2

Apply Newton’s method

f t( ) 0t = t −1 0 ƒ ′(t0 ) 0.5 − 2 = 0.5 −

e0.5 + 4e

= 0.5621…

Hence t1 = 0.56 is another approximate solution.

– 20 –


Question 14 (c) (ii)


• Uses part (c) (i) to find an approximate solution to , or equivalent

merit 2

• Establishes , or equivalent merit 1

We need to determine r and x so that

1rx rx e = log x and re = e x (common point and equal slope of tangents).

From the equal slopes

1rx e = rx

Setting rx = t and using part (i), we have an approximate solution

rx = 0.56

Since the curves intersect at x, rx 0.56 log x = e = ee

0.56 ex = e = 5.758… now rx = 0.56

0.56 ∴ r = 5.758…

= 0.0972

r = 0.097

– 21 –

Mathematics Extension 1 2013 HSC Examination Mapping Grid

Section I

Question Marks Content Syllabus outcomes

1 1 16.2 PE3

2 1 15.1 HE4

3 1 2.8 PE3

4 1 16.1 PE3

5 1 11.5 HE6

6 1 15.2 HE4

7 1 18.1 H5, PE3

8 1 5.7 H5, PE2

9 1 15.3 HE4

10 1 1.4E PE3

Section II


11 (a) 1 16.3 PE3

11 (b) 2 15.5 HE4

11 (c) 2 18.2 HE3

11 (d) (i) 2 8.8 P7, HE4

11 (d) (ii) 2 10.5 H6, HE4

11 (e) 1 13.4 H5, HE7

11 (f) 3 11.5, 15.5 HE4, HE6

11 (g) 2 8.8, 15.5 P7, HE4

12 (a) (i) 1 5.7, 5.9 PE2

12 (a) (ii) 2 5.7, 5.9 PE2

12 (b) 3 11.4, 13.6E H5, H8, HE7

12 (c) 3 14.2E HE3

12 (d) (i) 2 6.5 P4

12 (d) (ii) 1 8.5, 10.6 P4, H5

12 (d) (iii) 1 8.5 P6

12 (e) 2 14.4 HE3

13 (a) (i) 1 14.1 HE5

13 (a) (ii) 2 14.1 HE5

– 22 –

2013 HSC Mathematics Extension 1 Mapping Grid


13 (b) (i) 2 6.7E, 9.6 PE3

13 (b) (ii) 2 9.6 PE3, PE4

13 (c) (i) 2 14.3 HE3

13 (c) (ii) 1 14.3 HE3

13 (c) (iii) 2 5.7, 14.3 PE2, HE3

13 (d) 3 2.3, 2.9 P4, PE2, PE3, PE6

14 (a) (i) 1 1.4E PE3

14 (a) (ii) 3 7.4 HE2

14 (b) (i) 1 17.3 HE3

14 (b) (ii) 2 17.3 HE3

14 (b) (iii) 3 17.3 HE3

14 (c) (i) 2 16.4 HE7

14 (c) (ii) 3 8.6, 12.2, 12.4, 12.5 P6, H3, HE7

– 23 –

2013 HSC Mathematics Extension 1 Marking guidelines

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